Suppose is a number field that is Galois over
with
group
.
Fix a prime
lying over
.

(Note: The decomposition group is called the ``splitting group''
in Swinnerton-Dyer. Everybody I know calls it the decomposition
group, so we will too.)
Let
denote the residue class field of
.
In this section we will prove that there is a natural exact sequence

where
is the of
, and
. The most interesting part of the proof is
showing that the natural map
is surjective.
We will also discuss the structure of
and introduce
Frobenius elements, which play a crucial roll in understanding Galois
representations.

Recall that acts on the set of primes
lying
over . Thus the decomposition group is the stabilizer in
of
. The orbit-stabilizer theorem implies that
equals the orbit of
, which by Theorem 13.2.2
equals the number of primes lying over , so
.

**Lemma 14.1.2**
*
The decomposition subgroups
corresponding to primes
lying over a given are all conjugate in .*
*Proof*.
We have

if and only if

. Thus

if and only if

, so

. The lemma now follows
because, by Theorem

13.2.2,

acts transitively on the set of

lying over

.

The decomposition group is extremely useful because it allows us
to see the extension
as a tower of extensions, such that at
each step in the tower we understand well the splitting behavior
of the primes lying over . Now might be a good time to glance
ahead at Figure 14.1.2 on page .

We characterize the fixed field of
as follows.

*Proof*.
First suppose

, and note that by Galois theory

, and by Theorem

13.2.2, the group

acts transitively on the primes of

lying over

. One of
these primes is

, and

fixes

by definition, so there is
only one prime of

lying over

, i.e.,

does not
split in

. Conversely, if

is such that

does not split in

, then

fixes

(since it is the only
prime over

), so

, hence

.

Thus does not split in going from to --it does some
combination of ramifying and staying inert. To fill in more of
the picture, the following proposition asserts that splits
completely and does not ramify in
.

**Proposition 14.1.4**
*
Let for our fixed prime and Galois extension
.
Let
be for
and .
Then and
, i.e., does not ramify and splits
completely in . Also
and
.*
*Proof*.
As mentioned right after Definition

14.1.1, the
orbit-stabilizer theorem implies that

, and
by Galois theory

.
Thus

Now

and

, so
we must have

and

.
Since

and

,
the proposition follows.

**Subsections**
William Stein
2004-05-06